Is the Lie bracket with an orthonormal basis $0$ always?

Question

Let $(X_i)$ be an orthonormal basis and $(, )$ an inner product, then is the following true?

$( [X_i, X_j], X_i ) = 0$?

I found that $(X_iX_j - X_jX_i,X_i) = X_iX_jX_i - X_jX_i^2$ So looks like if there is commutativty or depending on the inner product, the sum is $0$.

For example, define $(X,Y) = tr(XY)$ then it seems the basis does not need to be orthonormal to be $0.$

Let's say $X_i$ are matrices.

Answer 1

Let $E_{ij}$ denote the $2\times 2$ matrix with a $1$ in the $(i,j)$ slot and a $0$ elsewhere. Then $E_{11}, E_{12}, E_{21}, E_{22}$ is clearly a basis for the space of all $2\times 2$ matrices (over whatever field you want.)

Then note that $[E_{11},E_{12}] = E_{12}$. Now, if $h$ denotes any inner product at all, then $h([E_{11},E_{12}], E_{12}) = h(E_{12}, E_{12}) > 0$.